Questions & Answers

Question

Answers

a) GP

b) HP

c) AP

d) Sp

Answer
Verified

So, we get a relation between a, b and c. Now, take x raise to the power on both sides. We have: ${{x}^{b}}={{x}^{\dfrac{a+b}{2}}}$. Solve the equation to find the relation between ${{x}^{a}},{{x}^{b}},{{x}^{c}}$.

As we know that, if three numbers are in Arithmetic Progression, the middle number can be written as the mean of the first and last number. Therefore, for the given numbers: a, b and c, we can write as:

$b=\dfrac{a+c}{2}......(1)$

Multiplying both sides with 2, we can write equation (1) as:

$2b=a+c......(2)$

So, we have a relation between a, b and c.

Now, take x raise to the power on both sides of equation (2), we get:

${{x}^{2b}}={{x}^{a+c}}......(3)$

As we know that:

$\left( \begin{align}

& {{x}^{2n}}={{\left( {{x}^{n}} \right)}^{2}} \\

& {{x}^{m+n}}={{x}^{m}}\times {{x}^{n}} \\

\end{align} \right)$

So, we can write equation (3) as:

${{\left( {{x}^{b}} \right)}^{2}}={{x}^{a}}\times {{x}^{c}}......(4)$

So, we get a relation between ${{x}^{a}},{{x}^{b}},{{x}^{c}}$. Now, try to compare the given relation with different progressions.

According to AP, the middle number can be written as the mean of the first and last number. But in equation (4) the middle number is the square root of multiplication of the first and last number. So, ${{x}^{a}},{{x}^{b}},{{x}^{c}}$ are not in AP.

Also, according to the GP, the middle number is the square root of multiplication of first and last number. So, ${{x}^{a}},{{x}^{b}},{{x}^{c}}$ are in GP.

According to HP, the harmonic mean of three numbers is reciprocal of mean of reciprocal of first and last number, i.e. $b=\dfrac{2}{\dfrac{1}{a}+\dfrac{1}{c}}$ . So, it gives a complicated relation between a, b and c. So, ${{x}^{a}},{{x}^{b}},{{x}^{c}}$ are not in HP.

Always try to get a relation between what is given and what is asked. It makes it easier to find the answer.